I get $\fbox{$(\Delta t)=0.15\,\mbox{s}$}$, and (quite likely) "yes".
Problem 2
I get $\fbox{$L=3.0\,\mbox{m}$}$.
Problem 3
\[ v = \sqrt{\frac{T}{\mu}}
= \sqrt{\frac {k\cdot(\Delta L)\cdot [L_0+(\Delta L)]}{m}} = f\lambda = f\cdot 2\,\cdot[L_0+(\Delta L)] \]
Solve for the fundamental frequency in terms of parameters $m$, $L_0$, and $k$; and the variable $(\Delta L)$,
\[ f = \frac{1}{2} \cdot \sqrt{\frac{k}{m}} \cdot \sqrt{\frac{(\Delta L)}{L_0 + (\Delta L)}} \]
See the formula above. When the Bungee cord is stretched, the tension increases AND the linear mass density (how many
kilograms per meter) decreases. The one that is stretched more (second one) should have the higher velocity.
I get the ratio of fundamental frequencies to be $\fbox{$f_1/f_2 = 1/\sqrt{1.5} = 0.82 $}$. The one that is stretched less has the lower fundamental
frequency.
Problem 1
I get $\fbox{$v_{\mbox{max}}=28\,\mbox{m/s}$}$. [Where does this maximum speed occur in the ride?]
Sketch the diagram.
I get $\fbox{$(5T)=62\,\mbox{sec}$}$.
This is a simple harmonic oscillator problem, not a constant acceleration problem.
Some relevant relations are
\[
y_{\mbox{max}}=A\qquad\qquad
v_{\mbox{max}}=A\,\omega\qquad\qquad
a_{\mbox{max}}=A\omega^2
\]
where $A$ is the amplitude and $y=-55\,\mbox{m}$ at the ground. The un-asked preliminary question
is, "What is $\omega$, the angular frequency, for this ride?"
Problem 2
Sketch the diagrams. (c) is "constructive" and (d) is "destructive".
Problem 3
I get $\fbox{$\lambda=1.14\,\mbox{m}$}$.
I get $\fbox{$v=342\,\mbox{m/s}$}$.
I get $\fbox{$\lambda=0.57\,\mbox{m}$}$.
I get $\fbox{$\lambda=5.10\,\mbox{m}$}$.
Problem 1
I get $\fbox{$(T/4=6.1\,\mbox{s}$}$. (Where is the $8^\circ$ part used in this problem?)
I get $\fbox{$(T/4=6.1\,\mbox{s}$}$. (Same as previous since it is still a small angle.)
Problem 2
I get $\fbox{$L{_\mbox{(Ohio)}}=0.9929\,\mbox{m}$}$.
I get $\fbox{$L{_\mbox{(Peru)}}=0.9893\,\mbox{m}$}$.
When both clocks are in Ohio, the shorter one (adjusted in Peru) will run
faster since it has a shorter period ($T=2\pi\sqrt{L/g}$).
Problem 1
velocities ranked: fastest 2 > 3 > 1 slowest
diameters ranked: largest 2 > 1 > 3 smallest
Problem 2
I get 3 minutes and 14 seconds.
Problem 3
I get about 51 grams for the candle (lantern). It should be lighter so the lantern will float up.
The upward normal forces on the beams are $N_1$ (left beam) and $N_2$ (right beam), and
$s$ is the distance to the right of the right beam that Ant Man is standing.
$\fbox{$N_1=0\,\mbox{N}$}$ (on verge of tipping)
$\fbox{$N_2=900\,\mbox{N}$}$ (supporting all weight above it, and 2 sig-figs $9.0\times 10^2\,\mbox{N}$)
$\fbox{$s=0.041\,\mbox{m}$}$
Problem 2
I get a net force of $\fbox{$F=360\,\mbox{N}$}$ downward (onto the jar).
Problem 3
The density of the oil is $\fbox{$\rho_{\mbox{oil}}=0.86\cdot\rho_{\mbox{H\textsubscript{2}}O}=0.86\,\mbox{gram}/\mbox{cm}^3=860\,\mbox{kg}/\mbox{m}^3$}$
Problem 1
The static equilibrium formula for torque should be
\[ (\Sigma\tau)_{\mathcal{O}} = (2.0)\cdot(T\,\sin(40^\circ))-((15)(9.8))(1.0)-((20)(9.8))(1.8) = 0 \]
where axis $\mathcal{0}$ is the left end of the beam.
This leads to the value for the tension to be $\fbox{$T=389\,\mbox{N}$}$.
The strategy is to use $(\Sigma F)_x=0$ and $(\Sigma F)_y=0$ to find the two components of the
unknown force that the wall exerts on the beam. We can actually do it to get:
\begin{align} & (\Sigma F)_x = F_x - T \,\cos(40^\circ)=0 \qquad\rightarrow\qquad F_x = 298\,\mbox{N} \\[2ex]
&(\Sigma F)_y = F_y + T\,\sin(40^\circ) - ((15)(9.8)) - ((20)(9.8)) = 0
\qquad\rightarrow\qquad F_y = 93\,\mbox{N}
\end{align}
Here is a to-scale free body digram for the beam:
Problem 1
I get a relatively simple formula that looks like:
\[ v = \sqrt{\frac{2g\,r\,\Delta \theta}{1+\tfrac12 M_p/m}} \]
It might be easier and quicker to solve it numerically in steps.
This gives a speed of $\fbox{$v=1.3\,\mbox{m/s}$}$.
(When you study this problem later, you can compare it to
$v=\sqrt{2gh}$ which we get for the speed of an object that is
dropped through a distance $h$.)
Yes. It is possible to get the answer in a similar number of steps, or more steps.
(What are those other physics prinicples?)
Problem 2
For an axially-symmetric object, find a symmetry axis where the mass is
as close to that axis as possible. This happens with (A) and (D).
What about the wheel?
Problem 3
I get $\fbox{$\omega_f=0.71\,\mbox{rad}/\mbox{s}$}$, the playground toy slows down.
I get $\fbox{$\Sigma\tau = -41\,mbox{N}\,\mbox{m}$}$.
Use $(\Sigma\tau) = (\Sigma F)\cdot R$ to get
\[ (\Sigma F) = \frac{(3.92)}{(0.50)} = 7.85\,\mbox{N} \]
The weight of an apple (the fruit, or the phone brand)
is somewhere between one and two Newtons. (My Android phone is 230 grams which is about
$2.3\,\mbox{N}$. Comparing the force above to a one-Newton apple, it is almost eight
times bigger than that. Is that reasonable?
Zero. Explain why. The weight of the stone is about 18,000
Newtons and its radius is half of one meter. How can the torque be
zero? (Being pedantic: The torque would be caused by the Earth via the
gravitational force, a.k.a. the stone's weight; torque is not exerted
by gravity.)
Problem 1
$\fbox{$t=62\,\mbox{s}$}$
It will affect the result, but the question becomes, with more
coins tossed, and less mass remaining, will it move faster or slower?
We can calculate it in the next part.
$\fbox{$t=41\,\mbox{s}$}$
For the over-ambitious, you can find a formula for the time it takes to be
\[ (\Delta t) =
\frac{(\Delta x)}{\sqrt{2\,m\,\mbox{KE}}} \cdot
\left[ M - m \right] =
\frac{(1.53\,\mbox{sec})}{\sqrt{m}}\cdot\left[(95\,\mbox{kg}) - m\right] \]
where $\mbox{KE}=85\,\mbox{J}$, $(\Delta x)=20\,\mbox{m}$, $M=95\,\mbox{kg}$
(total mass), and $m$ is the mass of coins that are thrown.
If $m=0$, it will take an infinite amount of time,
and if $m=30\,\mbox{kg}$ (all of the gold coins), it will take 18 seconds.
This more difficult problem is actually within the scope if this class.
As stated in the problem,
the kinetic energy is acquired only by the thrown coins. The above is the
correct, intended answer. However,
a more realistic situation is that a certain fixed amount of energy
is available for propelling him across the ice. That energy is divided into
both things that are moving, the thrown coins and Indiana with the remaining coins.
We would have to solve this problem instead,
\[ \mbox{KE} = \tfrac12 (M-m)u^2 + \tfrac12 mv^2 \qquad\mbox{and}\qquad
(M-m)u = mv \qquad\mbox{and}\qquad (\Delta t)=\frac{(\Delta x)}{u} \]
where $u$ is the speed of Indiana carrying the remaining coins.
The first equation is conservation of energy, the second one
is conservation of momentum, and the third is kinematics to find the time. I get
a similar looking expression,
\[ (\Delta t) =
\frac{(\Delta x)}{\sqrt{2\,m\,\mbox{KE}}} \cdot
\sqrt{ M\cdot \left[ M - m \right]} =
\frac{(1.53\,\mbox{sec})}{\sqrt{m}}\cdot\sqrt{(95\,\mbox{kg})\cdot\left[(95\,\mbox{kg}) - m\right]} \]
For the two masses, I get
\[ (\Delta t)(\mbox{with $m=5\,\mbox{kg}$}) = 63\,\mbox{s} \]
\[ (\Delta t)(\mbox{with $m=10\,\mbox{kg}$}) = 43\,\mbox{s} \]
Interestingly, when the thrown mass is very small (compared to the
total mass), the answer doesn't differ greatly from before. Even if
he throws the entire 30 kilograms, the time is 22 seconds, which is
only a little more than the 18 seconds from before.
Problem 2
I get the position of the $m_3=67\,\mbox{gm}$ hanging mass to be 25 cm from the $m_1=75\,\mbox{gm}$ hanging mass
and 20 cm from the $m_2=15\,\mbox{gm}$ hanging mass.
If you have trouble, a further hint is that you can validate the numbers using
\[ (75)(15) = (67)(10) + (15)(30) \]
with a two-digit precision.
In that, I assume that the pivot point is at the string attached to the ceiling above.
Problem 3
$\fbox{$\alpha=1130\,\mbox{s}^{-2}$}$ (This is positive, meaning it is speeding up.)
$\fbox{$\alpha=-168\,\mbox{s}^{-2}$}$ (I represent the slowing down motion using the minus sign.)
Note, the word radian is often not included since it is a
made-up placeholder unit. To see that, consider an angle $\theta=s/r$
of a circle where $s$ is arc length in meters and $r$ is the radius,
also in meters: $\theta$ is in radians which is really meter/meter=1 (no unit).
For example,
$\omega_0=800\,\mbox{RPM}=83.8\,\mbox{rad/s}=83.8\,\mbox{s}^{-1}$.
There are $2\pi$ radians in one revolution or cycle.
Problem 1
Even though it is not asked for in the first two parts, you should draw before and after diagrams, and
an $x\mbox{-axis}$.
$v_S'=28\,\mbox{m/s}$ (They ask for the speed. You should also be aware that this smaller puck
moves off to the right.)
$\mbox{KE}_i=75\,\mbox{J}\qquad\mbox{and}\qquad\mbox{KE}_f=8\,\mbox{J}+67\,\mbox{J}=75\,\mbox{J}$.
It is a perfectly elastic collision because $\mbox{KE}_i=\mbox{KE}_f$.
[make the drawing] The momenta are $p_S=3.57\,\mbox{kg}\,\mbox{m/s}$ to the right,
$p_S'=1.19\,\mbox{kg}\,\mbox{m/s}$
to the left, and $p_L'=2.38\,\mbox{kg}\,\mbox{m/s}$ to the right.
[the answers are indicated in the previous part.] $v_S'=7\,\mbox{m/s}$ to the left.
Problem 1
[show this] Hint, unit of power is Watt=Joule/second, and unit of Joule=Nm.
[show this] Hint, use definition of work and $v=(\Delta x)/(\Delta t)$.
I get $\bar{F}_{\mbox{resistive}}=265\,\mbox{N}$.
Problem 2
I get $v'=-0.4\,\mbox{m/s}$ (the minus sign means it is moving to the left).
I get $F_{\mbox{ave}}=10\,\mbox{kN}$.
Hint, I get a total initial kinetic energy of
$\mbox{KE}_i=7720\,\mbox{J}$, and a final value of
$\mbox{KE}_f=21\,\mbox{J}$. The question asks for the individual
kinetic energies.
Further question, what is the rate at which that kinetic energy
was converted to other forms of energy? The "Power" in the
collision? It happened in $\Delta t=0.10\,\mbox{s}$. I get
$P=77\,\mbox{kW}$. The followup question is where did all that
kinetic energy go?
Problem 3
With $m_A=2.0\,\mbox{kg}$ and $m_B=5.0\,\mbox{kg}$,
I get the final speeds, $v'_A=1.69\,\mbox{m/s}$ and
$v'_B=0.68\,\mbox{m/s}$. The primes indicate the final
situation, the initial speeds are both zero. Which
way are they moving?
Problem 1
$v=0.46\,\mbox{m/s}$
$x=+4.0\,\mbox{cm}$
Problem 2
Conservation of energy is the reason ... together with this
being a projectile motion problem with no effects due to the air
(like air resistance or buoyancy). As also shown in the next part, the speed of the water balloon
at two elevations is related by $v_1^2 = v_0^2 + 2gh$ where $v_0$
is the slower speed at height $h$ above where the faster speed is
$v_1$. It does not depend on the launch angle, which can be
anywhere from straight up to straight down.
Follow the equations, where Event 0 is when it is just after it is launched and Event 1 is just
before it hits the ground:
\begin{align}
KE_0 + PE_{g0} &= KE_1 + PE_{g1} \\[2ex]
\tfrac12\,m\,v_0^2 + mgy_0 &= \tfrac12\,m\,v_1^2 + mgy_1 \\[2ex]
v_0^2 + 2mgh &= v_1^2 + 2mg\cdot(0) \\[2ex]
v_0 = \sqrt{v_1^2 - 2mgh}
\end{align}
I get $\fbox{$v_0=45\,\mbox{m/s}$}$.
Problem 1
The static friction force causes Manuel to move in a
circle. (If there were suddenly no static friction, Manuel would move
in a straight line until he reaches the perimeter of the carousel.)
I get $R=2.0\,\mbox{m}$.
With the same static coefficent of friction (which is why all the fuss about the shoes),
that radius should not depend on the mass – within limits – of the person.
Problem 2
To get the time to go around once,
\[
f = \frac{1.5\,\mbox{rev}}{\mbox{min}}\times\frac{1\mbox{min}}{60\,\mbox{sec}}
= \frac{0.025\,\mbox{rev}}{\mbox{sec}} = \frac{1\,\mbox{rev}}{40\,\mbox{sec}} = \frac{1}{T}
\]
from which we get the period $T=40\,\mbox{s}$, which is the time for one revolution.
You should also get $v=3.96\,\mbox{m/s}$ and $mv^2/R=40\,\mbox{N}$.
$N_{\mbox{halfway}}=\sqrt{(637)^2+(40)^2} = 638\,\mbox{N}$ (mostly up, but
tipped $3.6^\circ$ toward the center)
The normal forces are not all equal. The normal force is computed
using \[ \Sigma\vec{F}=\vec{N}+\vec{W}=m\vec{a} \]
and with this
problem, $\vec{W}$ a constant 637-N force pointing straight down and
$\Sigma\vec{F}=m\vec{a}$ is the 40-N net force pointing toward the center
of the Ferris wheel.
Problem 3
Here, $\mbox{Work}_A$ is the work done by [the thing exerting] the force $F_A$.
$\mbox{Work}_A=280\,\mbox{J}$
$\mbox{Work}_A=243\,\mbox{J}$
$\mbox{Work}_A=280\,\mbox{J}$
$\mbox{Work}_A=280\,\mbox{J}$
$\mbox{Work}_A=243\,\mbox{J}$
$\mbox{Work}_A=280\,\mbox{J}$
To remove the ambiguity in this problem, assume that the box is initially not moving.
Being pedantic, forces don't do work. The thing that exerts that
force does the work, which is the amount by which the energy changes.
If I lift a heavy box, it is I who gets tired, not the normal force I
exert.
These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via
N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.
These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via
N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.
Problem 1
to the right, static friction
to the left, kinetic friction
the value of the static friction is zero. There is no slipping, so it is static friction, and the friction has no direction because it is zero. Note that it is not kinetic friction because the box is not sliding, even
though it is moving relative to the road (as the van drives down the road).
to the West, static friction
to the West, static friction
Problem 2
The desired answer is $\fbox{$d=54\,\mbox{m}$}$.
You should also be able to figure out the following:
$W=441\,\mbox{kN}$ (weight)
$N=400\,\mbox{kN}$ (normal force)
$F_k=240\,\mbox{kN}$ (kinetic friction)
$a = 9.5\,\mbox{m/s}^2$ (acceleration magnitude as it slows down)
In those, $1\,\mbox{kN}=1000\,\mbox{N}$ is a kiloNewton.
We should find the acceleration first.
Let down the ramp be the positive $x$ axis, and the positive $y$ axis be normal to the ramp and upward.
Newton's second law in component form then looks like:
\begin{align}
& (\Sigma F)_x = W \sin(\theta) + F_k = m a\\[2ex]
& (\Sigma F)_y = N - W \cos(\theta) = 0
\end{align}
Use the special formulas $W=mg$ and $F_k = \mu_k N$ to rewrite these as
\begin{align}
& mg\, \sin(\theta) + \mu_k N = m a\\[2ex]
& N = mg\,\cos(\theta)
\end{align}
Substitute the $N$ from the second equation into the first and solve for $a$ to get
\begin{align}
a = g\cdot\left[ \sin(\theta) + \mu_k\,\cos(\theta) \right]
\end{align}
Note how this does not depend on mass.
The next step is to use kinematics to find the distance. I find this formula works great:
\begin{align}
v_f^2 = v_i^2 + 2a D \\[2ex]
\rightarrow \quad d = |D| = \frac{v_i^2}{2a}
\end{align}
where $v_f=0$ and $D$ is negative because the $x$ axis points down the hill and the truck is moving up.
I use $d=|D|=|x_f-x_i|$ to be the distance the truck moves while stopping.
To get the numerical values of the forces, use the numbers in this order,
\begin{align}
& W = mg \\[2ex]
& N = W\,\cos(\theta) \\[2ex]
& F_k = \mu_k\,N
\end{align}
Problem 3
The correct answers I get for all the force magnitudes are
\begin{align}
& W = 12.7\,\mbox{N}\qquad\mbox{(weight of block)}\\[2ex]
& \fbox{$N = 4.5\,\mbox{N}$}\qquad\mbox{(normal force)} \\[2ex]
& T = 10.1\,\mbox{N}\qquad\mbox{(tension force)}\\[2ex]
& F_k = 2.7\,\mbox{N}\qquad\mbox{(kinetic friction force)}
\end{align}
The two equations I get are
\begin{align}
(\Sigma F)_x &= T \cos(\theta) - F_k = ma \\[2ex]
(\Sigma F)_y &= T \sin(\theta) +N - W =0
\end{align}
I use the two special formulas, $W=mg$ and $F_k=\mu_k N$, to rewrite them as
\begin{align}
& T\cos(\theta) -\mu_k N = ma \\[2ex]
& T \sin(\theta) + N = mg
\end{align}
There are two unknowns in these equations. They are $N$ and $T$. All of the other
quantities are given numerically. The value of $N$ is requested, but it is helpful
to find both unknowns. They are
\begin{align}
& N = \frac{m\left[ g - a \tan(\theta)\right]}{1+\mu_k \tan(\theta)} \\[2ex]
& T = \frac{m\left[ g\mu_k + a\right]}{\cos(\theta) + \mu_k \sin(\theta)} \\[2ex]
\end{align}
I can use maxima to get these:
$ maxima
(%i1) sol:solve([T*cos(t)-muk*N=m*a, N+T*sin(t)-m*g=0],[N,T]);
g m cos(t) - a m sin(t) g m muk + a m
(%o1) [[N = -----------------------, T = -------------------]]
muk sin(t) + cos(t) muk sin(t) + cos(t)
(%i2) float(subst([muk=0.60,m=1.3,t=55*%pi/180,g=9.8,a=2.4],sol));
(%o2) [[N = 4.461321647756916, T = 10.10640016094792]]
These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via
N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.
Problem 1
First look at the well-known Atwood machine (wikipedia.org) solution, and know
what it means. In that problem, there is only one string with one
tension. In the group work problem, there are two strings with two
tensions.
For the three-mass problem at hand, the solution I get is
$ a = 2.1\,\mbox{m/s}^2\qquad \mbox{(magnitude of acceleration of each mass)} $
$T_1 = 23.8\,\mbox{N}\qquad\mbox{(tension in string between A and B)}$
$T_2 = 38.5\,\mbox{N}\qquad\mbox{(tension in string between B and C)}$
There are various ways to express this answer. Here is one:
\begin{align}
19.6\,\mbox{N} = W_A \gt T_1 = 23.8\,\mbox{N} \\[2ex]
49.0\,\mbox{N} = W_C \lt T_2 = 38.5\,\mbox{N}
\end{align}
The intended point of this question is that the tension is not equal to the mass hanging on the string
if that hanging mass is accelerating.
Split it up into three separate problems, one for each of the three masses. You will get three independent
N2L equations with a total of three unknowns. Let $T_1$ be the tension in the string between A and B, and
$T_2$ be the tension in the string between B and C. Let $a$ be the magnitude of the acceleration
of each of the masses. The three equations are
\begin{align}
\Sigma F(\mbox{on A}) &= m_A\, a \qquad \Rightarrow & \qquad T_1 - W_A = m_A\, a \\[2ex]
\Sigma F(\mbox{on B}) &= m_B\, a \qquad \Rightarrow & \qquad T_2 - T_1 = m_B\, a \\[2ex]
\Sigma F(\mbox{on C}) &= m_C\, a \qquad \Rightarrow & \qquad W_C - T_2 = m_C\, a
\end{align}
In that, we only need the component of the force acting in the same direction as the respective acceleration,
which is why you should sketch a coordinate axis and motion diagram along with each free body diagram.
I also set up the three coordinate axes so that the "$ma$" term is positive for each.
Use the special formula for weight, $W=mg$, to get
\begin{align}
& T_1 - m_A\, g = m_A\, a \\[2ex]
& T_2 - T_1 = m_B\, a \\[2ex]
& m_C\, g - T_2 = m_C\, a
\end{align}
These are three [independent] equations with three unknowns, $a$, $T_1$, and $T_2$. The solution
is
\begin{align}
a &= g\cdot\left[\frac{m_C - m_A}{m_A+m_B+m_C} \right] \\[2ex]
T_1 &= m_A\cdot\left(g+a\right) \\[2ex]
T_2 &= m_C\cdot\left(g-a\right)
\end{align}
For those of you well-versed in the way of computer algebra systems, I offer this:
Maple:
> solve([ T1 - mA*g = mA*a, T2 - T1 = mB*a, mC*g - T2 = mC*a], [a,T1,T2]);
g (mA - mC) g mA (mB + 2 mC) mC g (2 mA + mB)
[[a = - ------------, T1 = ----------------, T2 = ----------------]]
mA + mB + mC mA + mB + mC mA + mB + mC
Maxima:
(%i1) solve([ T1 - mA*g = mA*a, T2 - T1 = mB*a, mC*g - T2 = mC*a], [a,T1,T2]);
g mA - g mC g mA (2 mC + mB) g mB mC + 2 g mA mC
(%o1) [[a = - ------------, T1 = ----------------, T2 = -------------------]]
mC + mB + mA mC + mB + mA mC + mB + mA
If you do it by hand, add them all together like this,
\begin{align}
& T_1 - m_A\, g = m_A\, a \\[1ex]
& T_2 - T_1 = m_B\, a \\[1ex]
& m_C\, g - T_2 = m_C\, a \\[0ex]
&\rule{2in}{0.5pt}\mbox{(add)} \\[1ex]
& m_C\,g - m_A\,g = m_A\,a+m_B\,a+m_C\,a \\[2ex]
& \left(m_C - m_A\right)\cdot g = \left(m_A+m_B+m_C\right)\cdot a
\end{align}
from which we can get $a$ (magnitude of acceleration). Next, use the first equation
to get $T_1 = m_A\cdot(g+a)$ and the third equation to get $T_2=m_C\cdot(g-a)$.
Do not do it this way until and unless you understand the way described above. It is easy to get it wrong.
A force of $W_A=m_A\,g$ pulls "one way (left)" and a force of $W_C=m_C\,g$ pulls "the other way (right)".
The total mass that is accelerated by these forces is $m_A+m_B+m_C$. Therefore, using $(\Sigma F)=ma$,
\begin{align}
W_C - W_A &= \left(m_A+m_B+m_C\right)\,a \\[2ex]
\left(m_C - m_A\right)\,g &= \left(m_A+m_B+m_C\right)\,a \\[2ex]
\end{align}
which gives the same expression for $a$. With $a$, it is straightforward to get the tensions, as above.
This method is correct; but, the misconception that may be
improperly reinforced is that the weight $W_A$ is pulling on B to the
left and the weight $W_C$ is pulling on B to the right. That is not
true since the strings are pulling on B, not the weights, and the
tensions are less than those weights (as Part C wants you to
recognize). The part that is often missed is that all three masses are accelerated
by that unbalanced force $(W_C-W_A)$, not just B.
These problems are not intuitive unless you are a Newtonian thinker who correctly links forces to motion via
N2L $(\Sigma\vec{F})=m\vec{a}$ and N3L $\vec{F}(\mbox{A on B})=-\vec{F}(\mbox{B on A})$.
Problem 1
Problem 2
Draw a motion diagram and a free body diagram for the person for these problems.
Reading greater than Weight
Reading less than Weight
Reading equals the Weight
Reading equals the Weight
$a=1.2\,\mbox{m/s}^2$ Up
$a=1.4\,\mbox{m/s}^2$ Down
Problem 1
I get 1 and 3 as not rejected based on a units check.
Remember that the units of
\[ F = ma \qquad\qquad\mbox{are}\qquad\qquad \mbox{N}=\mbox{kg}\,\mbox{m/s}^2 \]
Problem 2
$(\Sigma F) = 8.1\times 10^6\,\mbox{N} $
The acceleration is $a_x=0.43\,\mbox{m/s}^2$, which is the x-component to the right; whle $a_y=0.00\,\mbox{m/s}^2$ is the y-component.
\begin{align}
F_x(\mbox{Train on Passenger}) &= +33\,\mbox{N} \qquad \mbox{(to the right, called static friction force)}\\[2ex]
F_y(\mbox{Train on Passenger}) &= +755\,\mbox{N} \qquad \mbox{(straight up, called the normal force)}\\[2ex]
\end{align}
Newton's third law can be used:
\begin{align}
F_x(\mbox{Passenger on Train}) &= -33\,\mbox{N} \qquad \mbox{(to the left, called static friction force)}\\[2ex]
F_y(\mbox{Passenger on Train}) &= -755\,\mbox{N} \qquad \mbox{(straight down, called the normal force)}\\[2ex]
\end{align}
Problem 1
SB2 is the right one. (Explain why.)
Problem 2
I get
\[ v_{0x}=19.6\,\mbox{m/s} \qquad \mbox{and}\qquad v_{0y}=12.5\,\mbox{m/s} \]
This means
\[ v_0 = 23.2\,\mbox{m/s} \qquad \mbox{and}\qquad \theta = 32.5^\circ \]
The x-motion and y-motion are independent.
Problem 3
I get for the initial velocity (at the hands):
\[ v_{0x}=6.14\,\mbox{m/s} \qquad \mbox{and}\qquad v_{0y}=5.00\,\mbox{m/s} \]
This means
\[ v_0 = 7.92\,\mbox{m/s} \qquad \mbox{and}\qquad \theta_0 = 39.2^\circ \]
which is moving upward and to the right, away from the hands.
I get for the final velocity (at the hoop):
\[ v_{1x}=6.14\,\mbox{m/s} \qquad \mbox{and}\qquad v_{1y}=-1.86\,\mbox{m/s} \]
This means
\[ v_1 = 6.42\,\mbox{m/s} \qquad \mbox{and}\qquad \theta = -16.9^\circ \]
which is moving downward and to the right, into the basket.
These are some answers I got:
Problem 1
$C=B=A$
$C=B=A$
$C>B>A$
$C>B>A$
(none)
Problem 2
Here I also include two extra angles: $45^\circ$ (max range) and $90^\circ$ (straight up).
A
B
C
D
$\theta$
$H$
$R$
$T$
$30^\circ$
$30.4\,\mbox{m}$
$211\,\mbox{m}$
$8.1\,\mbox{s}$
$45^\circ$
$60.8\,\mbox{m}$
$243\,\mbox{m}$
$11.5\,\mbox{s}$
$60^\circ$
$91.2\,\mbox{m}$
$211\,\mbox{m}$
$14.0\,\mbox{s}$
$90^\circ$
$121.6\,\mbox{m}$
$0\,\mbox{m}$
$16.2\,\mbox{s}$
The one launched at $30^\circ$ hit first and the time difference is $5.9\,\mbox{s}$.
You should be able to sketch the diagrams for each of these angles.
You should confirm the other data that I give above in the table.
Another thing you can do is figure out how the potato will move on Earth where $g=9.8\,\mbox{m/s}^2$ (while
still disregarding effects due to the air, like wind resistance and buoyancy).
And one more thing you can do is to convert the $v_0=30\,\mbox{m/s}$ to MPH (miles/hour) to help
develop an intuition for how fast that potato is moving.
These are some answers I got:
Problem 1
1, 2, 3, and 5 to not pass the unit check.
Problem 2
The separate times are
\begin{align}
(\Delta t)_{BC} &= 0.938\,\mbox{s} \\[1ex]
(\Delta t)_{CA} &= 1.200\,\mbox{s}
\end{align}
The total time is
\[
(\Delta t)_{BA} = (0.938)+(1.200) = 2.138\,\mbox{s}
\]
For more practice, you can figure out the initial (at B) and final (at A) velocities,
\begin{align}
v_{B} &= 2.35\,\mbox{m/s}\qquad \mbox{up the incline} \\[1ex]
v_{A} &= 3.00\,\mbox{m/s}\qquad \mbox{down the incline}
\end{align}
Displacements are equal to $330\,\mbox{m}$, but distances are not equal:
Alice goes $330\,\mbox{m}$ and Bob goes $368\,\mbox{m}$.
I get the numerical answer of $s=3.02\,\mbox{m}$.
A much more complete numerical solution to check your work against is
Event 0
Event 1
Event 2
(Note)
$t_0=0$
$t_1=0.785\,\mbox{s}$
$t_2=0.971\,\mbox{s}$
$t_2-t_1=0.186\,\mbox{s}$
$y_0=0$
$y_1=3.02\,\mbox{m}$
$y_2=4.62\,\mbox{m}$
$y_2-y_1=1.60\,\mbox{m}$
$v_0=0$
$v_1=7.69\,\mbox{m/s}$
$v_2=9.52\,\mbox{m/s}$
In that, the $y\,\mbox{-axis}$ points down with an origin at the drop point.
Event 0 is the object begins to fall, Event 1 is it passes the
top of the window, and Event 2 is it passes the bottom of the window.
The constant acceleration, $a=9.80\,\mbox{m/s}^2$, points down. The
usefulness of pointing the $y\mbox{-axis}$ down is there are no negative signs.
If you need help getting started: Plot $y$ vs $t$ and
write down $y=\tfrac12 g t^2$. Then identify the events and the knowns and unknowns.
These are some final numerical answers I got for Problem 3:
The three non-zero accelerations are
\[ a = \begin{cases}
-5.09\,\mbox{m/s}^2 & 0.00\lt t\lt 3.00\,\mbox{s}\\[1ex]
-1.39\,\mbox{m/s}^2 & 8.00\lt t\lt 10.00\,\mbox{s}\\[1ex]
+9.65\,\mbox{m/s}^2 & 17.10\lt t\lt 17.39\,\mbox{s}
\end{cases}
\]
This assumes that the final abrupt stop in those forty centimeters
is taken into account. The acceleration from 3 to 8 seconds and
from 10 to 17.1 seconds is zero.
$22.9\,\mbox{m}$
$17.39\,\mbox{s}$
$0.00\,\mbox{m/s}$
$2.63\,\mbox{m/s}=9.49\,\mbox{KPH}$
No FERPA-protected information (student names, grades, etc.) appears on this website.
This website does not use cookies, but it uses a little javascript mostly to render formulas, and even less php.
jones.3 (R Jones) Last Modified 14:55:06 20-Apr-2022